Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $a = \dfrac{x^2 - 4x - 45}{-5x + 45} \times \dfrac{-3x - 12}{6x + 30} $
Explanation: First factor the quadratic. $a = \dfrac{(x + 5)(x - 9)}{-5x + 45} \times \dfrac{-3x - 12}{6x + 30} $ Then factor out any other terms. $a = \dfrac{(x + 5)(x - 9)}{-5(x - 9)} \times \dfrac{-3(x + 4)}{6(x + 5)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ (x + 5)(x - 9) \times -3(x + 4) } { -5(x - 9) \times 6(x + 5) } $ $a = \dfrac{ -3(x + 5)(x - 9)(x + 4)}{ -30(x - 9)(x + 5)} $ Notice that $(x - 9)$ and $(x + 5)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ -3\cancel{(x + 5)}(x - 9)(x + 4)}{ -30(x - 9)\cancel{(x + 5)}} $ We are dividing by $x + 5$ , so $x + 5 \neq 0$ Therefore, $x \neq -5$ $a = \dfrac{ -3\cancel{(x + 5)}\cancel{(x - 9)}(x + 4)}{ -30\cancel{(x - 9)}\cancel{(x + 5)}} $ We are dividing by $x - 9$ , so $x - 9 \neq 0$ Therefore, $x \neq 9$ $a = \dfrac{-3(x + 4)}{-30} $ $a = \dfrac{x + 4}{10} ; \space x \neq -5 ; \space x \neq 9 $